prove the basic propotionality theorem
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Answers
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Answer:
Statement :- If a line is drawn parallel to one side o triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
R. T. P :- AD/DB =AE/EC
Construction :- Draw ∆ ABC, DE is parallel to BC. Join B, E and C, D then draw DM perpendicular to AC and EN perpendicular toAB.
Proof :- Area of ∆ ADE = 1/2 × AD×EN
Area of ∆ B D E = 1/2 × BD ×EN
So, ar(∆ ADE) / ar(∆ BDE)
= 1/2×AD × EN / 1/2 × BD × EN
= AD /BD.........(1)
Again Area of ∆ ADE = 1/2 × AE × DM
Area of ∆ CDE = 1/2 × EC × DM
ar ( ∆ ADE) /ar ( ∆ CDE)
= 1/2 ×AE ×DM /1 /2 × EC×DM
=AE /EC......... (2)
We know that ∆BDE and ∆ CDE are on the same base DE and between same parallels BC and DE.
So ar(∆BDE) = ar (∆ CDE).......... (3)
From (1), (2) and (3), we have
AD/DB = AE/EC
Hence proved....
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